Question

A circle is drawn with its centre of the line $(4,5)$ and pass through point $(0,1)$. Find its equation.

Answer 1

Equation of circle center at $(4,5)$ is given as $(x-4{)}^2+(y-5{)}^2=r^2$

It passes through $(0,1)$

$⟹4^2+(1-5{)}^2=r^{2}16+16=r^2{r}^2=32$

So the equation is $(x-4{)}^2+(y-5{)}^2=32x^2+y^2-8x-10y+9=0$